A334765 Numbers m such that the numbers of 1's in the binary expansion of m equals the negative sum of balanced ternary trits of m.

0, 41, 68, 131, 132, 368, 384, 528, 1095, 1098, 1100, 1106, 1112, 1122, 1124, 1152, 1176, 1346, 1824, 2561, 3282, 3284, 3336, 3344, 3392, 3524, 4098, 4101, 4104, 4112, 4118, 4128, 4172, 4352, 4496, 4739, 4740, 5504, 6224, 9856, 9857, 9869, 9896, 9923, 9924

Offset

1,2

Comments

a(116) = 32770, a(117) = 32771 and a(118) = 32772 is the first time that three consecutive numbers appear in this sequence. Conjecture: There is no occurrence of four or more consecutive numbers. Tested by exhaustive search up to a(n) = 3^26. - Thomas König, Jul 19 2020

Links

Thomas König, Table of n, a(n) for n = 1..30000

Formula

Integers m such that -A065363(m) = A000120(m).

Example

41_10 = 1TTTT_bt = 101001_2, the sum of the digits is 1-1-1-1-1 = -3 for balanced ternary and 1+1+1 = 3 for base 2, so 41 is a term.

Mathematica

Select[Range[0, 10^4], -Total@ If[First@ # == 0, Rest@ #, #] &[Prepend[IntegerDigits[#, 3], 0] //. {x___, y_, k_ /; k > 1, z___} :> {x, y + 1, -1, z}] == DigitCount[#, 2, 1] &] (* Michael De Vlieger, Jul 08 2020 *)

Prog

(PARI) bt(n)= if (n==0, return (0)); my(d=digits(n, 3), c=1); while(c, if(d[1]==2, d=concat(0, d)); c=0; for(i=2, #d, if(d[i]==2, d[i]=-1; d[i-1]+=1; c=1))); vecsum(d); \\ A065363
isok(m) = bt(m) + hammingweight(m) == 0; \\ Michel Marcus, Jun 07 2020

Crossrefs

Aside from the first term, subsequence of A174657.

Cf. A000120, A037301, A065363, A330904.

Sequence in context: A039524 A259586 A140374 * A269807 A289982 A054806

Adjacent sequences: A334762 A334763 A334764 * A334766 A334767 A334768

Keyword

base,nonn

Author

Thomas König, May 10 2020

Status

approved